# UFR / UF Coset Space

Following up on an earlier thread I have explored the UFR/UF coset space. The number of UF cosets in the UFR group is given by:

(7 x 3) x (9 x 2) x (8 x 2) x 6 x 26 = 2,322,432

The first three factors are the number of ways the corner cubie position and the two edge cubie positions which are not on the UF faces may be configured. The factor of 6 is for the corner position permutation. Of the 720 corner position permutations on the UF faces achievable using the UFR face turns only 120 are achievable using the UF face turns. Thus the corner permutation may be one of six cosets. The flip of the seven UF edge cubies is constrained under the UF face turns, one cannot perform a double edge flip in this group. This gives rise to a factor of two to the sixth for the edge flip (flip parity determines the flip of the seventh edge cubie). As Bruce Norskrog pointed out in the earlier thread, the above number is the same as the order of the UFR group divided by the order of the UF group, so things check out.

Using the above analysis to uniquely designate which of the 2,322,432 cosets an element of the UFR group belongs to I have characterized the elements of UFR group by coset to a depth of 15 q turns:

 Depth Cosets Subtotal 0 1 1 1 2 3 2 9 12 3 40 52 4 172 224 5 725 949 6 2,972 3,921 7 11,991 15,912 8 46,117 62,029 9 157,775 219,804 10 436,771 656,575 11 799,123 1,455,698 12 688,908 2,144,606 13 172,118 2,316,724 14 5,702 2,322,426 15 6 2,322,432

This shows that starting from an arbitrary member of the UFR group a member of the UF group may be found in a most 15 q turns. Therefore the worst case for my solution algorithm which proceeds < Rubik Group > --> < UFR Group > --> < UF Group > --> Identity is 49 q turns. Nine turns to solve a 2 x 2 corner, 15 turns to find a member of the UF group, and 25 turns to solve the remaining two faces.

In a related matter, in an attempt to discover an elegant method to determine which coset a corner permutation belongs to I performed a conjugate class characterization of the elements of the S6 group and the UF corner permutations:

 Class Cycle Structure S6 UF UF Coset 1 (1)(1)(1)(1)(1)(1) 1 1 0 2 (2)(1)(1)(1)(1) 15 0 3 3 (3)(1)(1)(1) 40 0 8 4 (2)(2)(1)(1) 45 15 6 5 (4)(1)(1) 90 30 12 6 (3)(2)(1) 120 0 24 7 (5)(1) 144 24 24 8 (2)(2)(2) 15 10 1 9 (3)(3) 40 20 4 10 (4)(2) 90 0 18 11 (6) 120 20 20 Totals 720 120 120

This attempt was ultimately a failure. I couldn't see how to use this analysis to characterize each coset. The patterns are intriguing. Why in class 4 is a crosswise double swap on the up face an element of the UF group but a double swap in parallel is not? I couldn't figure it out so I ended up making a table of all 720 permutations sorted by coset in order to assign a coset number to an arbitrary permutation. Anyway, I thought the above might be of interest to someone.